用c语言编写的浮点数的范围？ [英] Range of magnitude for float in c programming language?

问题描述

在丹尼斯·里奇（Dennis Ritchie）的C编程语言一书中提到，浮点数通常是一个32位数，至少有六位有效数字，幅度一般在10 ^ -38到10 ^ +38之间。

由于我们只有32位，这怎么可能呢？上限不应该是2 ^ 32？我试着打印出一个浮点数，循环乘以浮点数10,38次，这就是输出1000000069440617300000000000000000000.0000000000。它也必须跟踪这个符号，那么它如何将所有这些存储在32位？解析方案

参见< a href =https://en.wikipedia.org/wiki/Single-precision_floating-point_format =nofollow noreferrer>单精度浮点格式，了解关于典型C 浮点。

c语言中浮点数的范围？

  #include< float.h> 
 printf（浮动幅度范围％e到％e \ n，FLT_MIN，FLT_MAX）; 
 //典型结果
 //浮动幅度范围1.175494e-38至3.402823e + 38 
  

这是可能的，因为我们只有32位？典型的浮动确实只能存储大约2 <32> 32

sup>不同的值。然而，它们不是线性分布的，而是以线性组对数形式的。

<2> 23 在范围[2 <-126 < sup>〜2 ^-127 ）

...

2 ²³ 范围[0.5到1.0）

2 ²³ 在范围[1.0至2.0]中的不同值 -
2 ²³ 4.0）

...

2 ²³ 范围内的不同值[2 127到2 128 ）

以及它们的负值计数器部分。
也+/- 0零，小的子正常数字，+/-无穷大而不是-a-Number

它还必须跟踪符号，那么它如何将所有这些存储在32位？ / b>

  1位为符号
 8位为二进制指数
 23对于重要的位（MSBit通常表示为1）
  -  
 32位
  

---

何时打印一个 float ，只有大约6个（ FLT_DIG，FLT_DECIMAL_DIG ）有效数字通常很重要。

  printf（％。* e \\\
，FLT_DIG-1，FLT_TRUE_MIN）; 
 printf（％。* e \\\
，FLT_DIG-1，FLT_MIN）; 
 printf（％。* e \\\
，FLT_DIG-1，acos（-1））; 
 printf（％。* e \\\
，FLT_DECIMAL_DIG  -  1，nextafterf（FLT_MAX，0.0））; 
 printf（％。* e \\\
，FLT_DECIMAL_DIG  -  1，FLT_MAX）; 
  

输出

  1.40130e-45 // min sub-normal 
 1.17549e-38 // min normal normal 
 3.14159e + 00 // pi 
 3.40282326e + 38 //最大值之前的数字
 3.40282347e + 38 //最大
  

In The C Programming Language book by Dennis Ritchie, it is mentioned that "A float number is typically a 32-bit quantity, with at least six significant digits and magnitude generally between about 10^-38 and 10^+38."

How is that possible since we have only 32-bits? Shouldn't the top limit be 2^32? I tried printing out a float by looping and multiplying a float by 10, 38 times and this was the output 100000006944061730000000000000000000000.000000 . It also has to keep track of the sign , so how is it storing all this in just 32-bits?

解决方案

See Single-precision floating-point format for a details about a typical C float.

Range of magnitude for float in c programming language?

#include <float.h>
printf("float magnitude range %e to %e\n", FLT_MIN, FLT_MAX);
// typical result
// float magnitude range 1.175494e-38 to 3.402823e+38

How is that possible since we have only 32-bits?

Typical float does indeed only store about 2³² different values. Yet they are not distributed linearly but logarithmically in linear groups.

2²³ different values in the range [2^-126 to 2^-127)
...
2²³ different values in the range [0.5 to 1.0)
2²³ different values in the range [1.0 to 2.0)
2²³ different values in the range [2.0 to 4.0)
...
2²³ different values in the range [2¹²⁷ to 2¹²⁸)

And their negative counter parts.
Also +/- zeros, small sub-normal numbers, +/- infinity and Not-a-Number

It also has to keep track of the sign , so how is it storing all this in just 32-bits?

 1 bit for sign
 8 bits for the binary exponent
23 bits for the significant (with a MSBit usually implied as 1)
--
32 bits

---

When printing a float, only about 6 or so (FLT_DIG, FLT_DECIMAL_DIG) significant digits usually are important.

printf("%.*e\n", FLT_DIG-1, FLT_TRUE_MIN);
printf("%.*e\n", FLT_DIG-1, FLT_MIN);
printf("%.*e\n", FLT_DIG-1, acos(-1));
printf("%.*e\n", FLT_DECIMAL_DIG - 1, nextafterf(FLT_MAX, 0.0));
printf("%.*e\n", FLT_DECIMAL_DIG - 1, FLT_MAX);

Output

1.40130e-45    // min sub-normal
1.17549e-38    // min normal
3.14159e+00    // pi
3.40282326e+38 // number before max
3.40282347e+38 // max

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